$ \left(\dfrac{125}{8}\right)^{-\frac{4}{3}}$
Solution: $= \left(\dfrac{8}{125}\right)^{\frac{4}{3}}$ $= \left(\left(\dfrac{8}{125}\right)^{\frac{1}{3}}\right)^{4}$ To simplify $\left(\dfrac{8}{125}\right)^{\frac{1}{3}}$ , figure out what goes in the blank: $\left(? \right)^{3}=\dfrac{8}{125}$ To simplify $\left(\dfrac{8}{125}\right)^{\frac{1}{3}}$ , figure out what goes in the blank: $\left({\dfrac{2}{5}}\right)^{3}=\dfrac{8}{125}$ so $ \left(\dfrac{8}{125}\right)^{\frac{1}{3}}=\dfrac{2}{5}$ So $\left(\dfrac{8}{125}\right)^{\frac{4}{3}}=\left(\left(\dfrac{8}{125}\right)^{\frac{1}{3}}\right)^{4}=\left(\dfrac{2}{5}\right)^{4}$ $= \left(\dfrac{2}{5}\right)\cdot\left(\dfrac{2}{5}\right)\cdot \left(\dfrac{2}{5}\right)\cdot \left(\dfrac{2}{5}\right)$ $= \dfrac{4}{25}\cdot\left(\dfrac{2}{5}\right)\cdot \left(\dfrac{2}{5}\right)$ $= \dfrac{8}{125}\cdot\left(\dfrac{2}{5}\right)$ $= \dfrac{16}{625}$